When sulphur dioxide is passed in an acidified `K_(2)Cr_(2)O_(7)` solution, the oxidation state of sulphur is changed from:

Options

A. `+4` to 0
B. `+4 `to 2
C. `+4` to +6
D. `+6` to +4

Explanation

To determine how the oxidation state of sulfur changes when sulfur dioxide (SO₂) is passed through an acidified potassium dichromate (K₂Cr₂O₇) solution, we can follow these steps:

### Step 1: Identify the reactants and products
When sulfur dioxide reacts with acidified potassium dichromate, the reaction can be represented as:
\[ \text{3 SO}_2 + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{3 SO}_4^{2-} + \text{2 Cr}^{3+} + \text{H}_2\text{O} \]

### Step 2: Determine the oxidation state of sulfur in SO₂
In sulfur dioxide (SO₂), the oxidation state of sulfur (S) can be calculated as follows:
- Oxygen (O) has an oxidation state of -2.
- The formula for SO₂ is: S + 2(-2) = 0 (since the molecule is neutral).
- Therefore, S - 4 = 0 → S = +4.

### Step 3: Determine the oxidation state of sulfur in SO₄²⁻
In sulfate ion (SO₄²⁻), the oxidation state of sulfur can be calculated similarly:
- The sulfate ion has a charge of -2.
- The formula for SO₄²⁻ is: S + 4(-2) = -2.
- Therefore, S - 8 = -2 → S = +6.

### Step 4: Compare the oxidation states
Now we can compare the oxidation states of sulfur in the two compounds:
- In SO₂, the oxidation state of sulfur is +4.
- In SO₄²⁻, the oxidation state of sulfur is +6.

### Conclusion
The oxidation state of sulfur changes from +4 in SO₂ to +6 in SO₄²⁻ when it reacts with acidified K₂Cr₂O₇.

### Final Answer
The oxidation state of sulfur changes from +4 to +6.

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